[leetcode]First Bad Version

产品团队在开发新产品过程中遇到质量检查失败的问题。为了找到第一个导致后续版本不良的版本,实现了一个函数来最小化调用API次数。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad version are also bad.

Suppose you have n versions [1, 2, ..., n] and you want to find out the first bad one, which causes all the following ones to be bad.

You are given an API bool isBadVersion(version) which will return whether version is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.

/* The isBadVersion API is defined in the parent class VersionControl.
      boolean isBadVersion(int version); */

public class Solution extends VersionControl {
    public int firstBadVersion(int n) {
        long st = 1, ed = n;
        Solution v = new Solution();
        while(st < ed) {
            int mid = (int)((st+ed)>>1);
            System.out.println(mid);
            boolean is = v.isBadVersion(mid);
            if(!is && v.isBadVersion(mid+1)) {
                return mid+1;
            } else if(is) {
                // x x
                ed = mid;
            } else {
                // v v 
                st = mid+1;
            }
        }
        return (int)st;
    }
}


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值